# Tagged: conjugate transpose

## Problem 336

A complex square ($n\times n$) matrix $A$ is called normal if
$A^* A=A A^*,$ where $A^*$ denotes the conjugate transpose of $A$, that is $A^*=\bar{A}^{\trans}$.
A matrix $A$ is said to be nilpotent if there exists a positive integer $k$ such that $A^k$ is the zero matrix.

(a) Prove that if $A$ is both normal and nilpotent, then $A$ is the zero matrix.
You may use the fact that every normal matrix is diagonalizable.

(b) Give a proof of (a) without referring to eigenvalues and diagonalization.

(c) Let $A, B$ be $n\times n$ complex matrices. Prove that if $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$, where $I$ is the $n\times n$ identity matrix.